Match List-I with List-II
| List-I | List-II | ||
| Function | Critical point | ||
| (A) | \( f(x)=x^3-3x \) | (I) | \( x=3 \) |
| (B) | \( f(x)=x^3-12x \) | (II) | \( x=0 \) |
| (C) | \( f(x)=x^2-6x \) | (III) | \( x=1 \) |
| (D) | \( f(x)=x^2 \) | (IV) | \( x=2 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
A critical point is a value of x where f′(x) = 0.
(A) \( f(x)=x^3-3x \)
This matches (III) \( x=1 \).
\( \displaystyle f'(x)=3x^2-3=0 \)
\( \displaystyle x=\pm 1 \)
(B) \( f(x)=x^3-12x \)
This matches (IV) \( x=2 \).
\( \displaystyle f'(x)=3x^2-12=0 \)
\( \displaystyle x=\pm 2 \)
(C) \( f(x)=x^2-6x \)
This matches (I) \( x=3 \).
\( \displaystyle f'(x)=2x-6=0 \)
\( \displaystyle x=3 \)
(D) \( f(x)=x^2 \)
This matches (II) \( x=0 \).
\( \displaystyle f'(x)=2x=0 \)
\( \displaystyle x=0 \)
(A) and (B) each have two critical points; List-II gives the positive one.
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).